JEE Physics 2025 Question with Solution

Given: The graph is a circle in the $PV$ diagram for a cyclic process.

Find: Work done by the gas in the entire cycle.

For a cyclic process, the work done by the gas is equal to the area enclosed by the cycle.

From the graph:

• Radius in pressure is $R_P = \frac{500 - 300}{2} = 100 \, \text{kPa}$
• Radius in volume is $R_V = \frac{350 - 150}{2} = 100 \, \text{cm}^3$

Using the area of the circular loop,

$$\text{Area} = \pi R^2 = 3.14 \times 100^2 = 3.14 \times 10^4$$

with units $\text{kPa} \cdot \text{cm}^3$.

Now convert units:

$$1 \, \text{kPa} \cdot \text{cm}^3 = 10^{-2} \, \text{J}$$

So,

$$W = 3.14 \times 10^4 \times 10^{-2} = 314 \, \text{J}$$

Hence, in the asked form,

$$W = 31.4 \times 10^{-1} \text{ J}$$

Therefore, the required numerical value is $31.4$.

Given: The diameters of the circular path on the axes are read directly from the graph.

Find: The work done enclosed by the loop.

The enclosed area can also be written as

$$W = \frac{\pi}{4} d_1 d_2$$

where

$$d_1 = (500 - 300) \times 10^3 \, \text{Pa}$$

and

$$d_2 = (350 - 150) \times 10^{-6} \, \text{m}^3$$

Substituting,

$$W = \frac{\pi}{4} (500 - 300) \times 10^3 \times (350 - 150) \times 10^{-6}$$ $$W = \frac{3.14}{4} \times 200 \times 10^3 \times 200 \times 10^{-6}$$ $$W = 31.4 \, \text{J}$$

Thus the work done is $31.4 \, \text{J}$.

Since the question asks for the value in the form _____ $\times 10^{-1}$ J, the required entry is $31.4$.